Eng. Math. Lett. 2018, 2018:2 https://doi.org/10.28919/eml/3820 ISSN: 2049-9337
LARGE OPEN FILTER BASES FOR FIXED POINTS
C. GANESA MOORTHY1AND S. IRUTHAYA RAJ2,∗
1Department of Mathematics, Alagappa University, Karaikudi-630 004, Tamil Nadu, India
2Department of Mathematics, Loyola College (Autonomous), Chennai-600034, Tamil Nadu, India
Copyright c2018 Moorthy and Raj. This is an open access article distributed under the Creative Commons Attribution License, which permits
unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract.A local base of the neighbourhood system associated with a fixed point is called a fixed filter base. The possibilities of constructing fixed filter bases are discussed, which lead to a concept of fixed point at infinity.
Keywords:filter base; fixed point; bounded set; topological vector space.
2010 AMS Subject Classification:47H10, 54H25.
1.
Introduction
A point in a metric space or a topological space is identified with its neighbourhood system
when topological extensions are studied (see, for example [2]). So, a fixed point of a mapping
may also be considered as a local base for the neighbourhood system. It is a filter base of
nonempty open sets containing the fixed point. If a local base for a neighbourhood system of a
point in a metric space is considered, then the diameter of the members of the local base tends
∗Corresponding author
E-mail addresses: [email protected] Received April 6, 2018
to zero, when the local base is considered as a directed set under the inclusion relation. The
following convention is to be followed in the first two sections of this article.
Convention 1.1. A filter baseU of nonempty open subsets of a metric space(X,d)is called a diametrically zero converging open filter base (or, DZCO-filter base) if the net(diam U)U∈U
of nonnegative extended real numbers converges to zero over the directed set U under the
inclusion relation. Here diam U =sup{d(x,y):x,y∈U}.
Thus,U is aDZCO-filter base in a metric space(X,d), if
(i) U ∈U =⇒U6= /0 andU is open;
(ii) U,V ∈U =⇒there existsW ∈U such thatW ⊆U∩V; and
(iii) diam U →0 asU varies in the directed setU.
Definition 1.2. Let f :X →X be a given mapping on a metric space (X,d). A DZCO-filter baseU in X is called a fixed filter base of f in X , if for given U ∈U, there is a V ∈U such
that f(V)⊆U .
If f :X →X is a mapping on a metric space (X,d), x ∈X is a fixed point of f, and f
is continuous at x, then U =B x,1n:n=1,2, . . . is a fixed filter base of f in X, when
B x,1n
=
y∈X :d(x,y)<1n . Continuity is essential to ensure that existence of a fixed point implies existence of a fixed filter base.
Example 1.3.
Define f :R→Rby f(x) =1+xforx6=0 and f(0) =0, whenRis the real line endowed with
the usual metric. This function f has the unique fixed point zero at which f is not continuous.
But there is no fixed filter base of f inR.
On the other hand the existence of a fixed filter base need not imply the existence of a fixed
point.
Define f :(0,1)→(0,1)by f(x) = 2x, forx∈(0,1)with the usual metric on(0,1). Then f
has a fixed filter baseU = 0,1n:n=1,2, . . . . This function f has no fixed point in(0,1). But a fixed point may be realized in the completion of(0,1).
Proposition 1.5. Let f :(X,d)→(X,d)be a function with a fixed filter baseU in a complete metric space (X,d). Suppose to each U ∈U, there is a V ∈U such that V , the closure of
V , is contained in U . Then there is a fixed point x∈X such that U is a local base for the
neighbourhood system for x.
Proof. Since(X,d)is complete and diam U →0 asU varies in U, then T U∈U
U ={x}for
somex∈X. By our hypothesis, the equality T U∈U
U={x}is also true. To eachU ∈U, there is
aV ∈U such that f(V)⊆U. Thus f(x)∈U,∀U∈U. This proves thatxis a fixed point of f.
Sincediam U →0 asU varies inU,x∈U andU is open for everyU ∈U, and T U∈U
U={x},
then the filter baseU should be a local base to the neighbourhood system forx.
Proposition 1.6. Let f :(X,d)→(X,d)be a continuous function with a fixed filter baseU in a compact metric space(X,d). Then there is a fixed point x∈X such that x∈U,∀U∈U.
Proof. Note that T U∈U
U ={x}, for somex∈X. To eachU ∈U, there is aV ∈U such that
f(V)⊆U; and by the continuity of f, f(x)∈ f(V)⊆ f(V)⊆Uare true. Thus f(x)∈ T U∈U
U=
{x}so that f(x) =x. This completes the proof.
Example 1.7.
Consider the continuous function f :[0,1]→[0,1] defined by f(x) = 12x, ∀x∈[0,1], when
[0,1] is endowed with the usual metric. The filter base U =(0,1n):n=1,2, . . . is a fixed filter base of f and 0 is a fixed point of f such that 0 is in the closure of(0,1n), for eachn. But
(0,1n)is not a neighbourhood of 0, for anyn.
Convention 1.8. A DZCO-filter baseU in a metric space is called a DZCOR-filter base, if for each U ∈U, there is a V ∈U such that V ∈U .
The filter base 0,1n
:n=1,2, . . . and0,1n
:n=1,2, . . . are fixed filter bases for the
function given in example 1.7. The first one is not a fixedR-filter base, whereas the second one
is a fixedR-filter base.
2.
Construction of fixed filter bases
Theorem 2.1. Let f :X →X be a mapping on a metric space(X,d)such that d(f(x),f(y))≤
kd(x,y)∀x,y∈X and for some fixed k∈(0,1). If x1∈X is fixed, and xn+1= f(xn), n=1,2, . . ., are defined, then
S
i≥n
B(xi,kn):n=1,2, . . .
is a fixed base of f in X .
Proof. Recall that B(x,r) ={y∈X:d(x,y)<r}. Write Bn,r = S i≥n
B(xi,r), when xn+1=
f(xn),n=1,2, . . ., andx1is fixed. The inequalities
d(xn+2,xn+1)≤kd(xn+1,xn)≤k2d(xn,xn−1)≤ · · · ≤knd(x2,x1)and
d(xn,xm)≤d(xn,xn+1) +d(xn+1,xn+2) +· · ·+d(xm−1,xm)
for m>n imply that lim
n→∞
diam{xn,xn+1, . . .}=0, and hence 0≤lim sup
n→∞
diam Bn,r ≤2r. If
x∈Bn,r thenx∈B(xi,r) for some i≥n, and henced(xi+1,f(x)) =d(f(xi),f(x))≤kd(xi,x)
imply that f(x)∈B(xi+1,kr). So,
f(Bn+1,kr)⊆ f(Bn+1,r)⊆ f(Bn,r)⊆Bn+1,kr⊆Bn+1,r⊆Bn,r.
Thus,Bn,kn :n=1,2, . . . is a filter base of nonempty open sets such that
(i) lim
n→∞
diam Bn,kn =0, and
(ii) f(Bn+1,kn+1)⊆Bn+1,kn+1 ⊆Bn,kn, for everyn=1,2, . . ..
So,{Bn,kn :n=1,2, . . .}is a fixed filter base of f. This completes the proof.
One may find a continuous extension of f to the completion of f, a fixed point of this
exten-sion as the limit of the sequencexnas in Banach contraction theorem, and thereby a fixedR-filter
base of f. When the construction of completion through Cauchy sequences (see comments
fol-lowing Proposition 2.1 in [1]) and the Banach fixed point iteration procedure (see Theorem 3.1
Theorem 2.2. Let f,X,d,k, and x1,x2, . . . be as in theorem2.1. Suppose further that(X,d)is complete. Then
nn
x∈X : lim
n→∞
d(xn,x)< m1o:m=1,2, . . .o is a fixed R-filter base of f in X .
These illustrated constructions reveal the difficulties in constructing fixed filter bases and
their dependence on the construction of fixed points. It is expected that some good methods
will be developed.
3.
Fixed point at infinity
The previous section gives a motivation to propose the next definition.
Definition 3.1. Let X be a locally compact Hausdorff space which is not compact. Let f :X→X be a mapping. Then f is said to have a fixed point at infinity, if for each compact subset K1of X , there is a compact subset K2of X such that f(X\K2)⊆X\K1.
Theorem 3.2. Let X and f be as in the definition3.1. Then f has a fixed point at infinity if and only if the closure of inverse image of any compact subset of X is compact.
Proof.Suppose f has a fixed point at infinity. LetK1be a compact subset ofX. Then there is
a compact subsetK2ofX such that f(X\K2)⊆X\K1. So, f−1(K1)⊆K2, and hence the closure of f−1(K1)is compact, becauseK2is compact.
Conversely assume that the closure of inverse image of any compact subset ofX is compact.
LetK1 be a compact subset ofX. LetK2be the closure of f−1(K1)inX. ThenK2 is compact
and f(X\K2)⊆X\K1.
Corollary 3.3. A homeomorphism from a locally compact Hausdorff space that is not compact onto itself has a fixed point at infinity.
The definition 3.1 agrees with the classical definition of point at infinity for one-point
com-pactification of a locally compact Hausdorff space that is not compact. However, it is possible
to extend the definition 3.1 and the theorem 3.2 to metric spaces in the following manner.
Definition 3.4. Let X be a metric space which is not bounded. Let f :X →X be a mapping. Then f is said to have a fixed point at infinity, if for every closed and bounded set C1in X , there is a closed and bounded set C2in X such that f(X\C2)⊆X\C1.
Theorem 3.5. Let X and f be as in the definition3.4. Then f has a fixed point at infinity, if and only if inverse image of any bounded subset of X is bounded in X .
Proof. Suppose f has a fixed point at infinity. LetC1be the closure of a given bounded set
BinX. ThenC1is also a bounded subset ofX. Then, there is a closed and bounded setC2in
X such that f(X\C2)⊆X\C1. Then f−1(B)⊆ f−1(C1)⊆C2, and hence f−1(B) is bounded, becauseC2is bounded.
Conversely, assume that inverse image of any bounded set is a bounded set inX. LetC1be a closed and bounded subset of X. LetC2 be the closure of f−1(C1). ThenC2 is closed and bounded and f(X\C2)⊆X\C1.
Corollary 3.6. Let f :(X,d)→(X,d)be a mapping on an unbounded metric space(X,d)such that
d(x,y)≤kd(f(x),f(y)),∀x,y∈X,
for some k>0. Then f has a fixed point at infinity.
Remark 3.7. Since bounded sets are also defined in Hausdorff topological vector spaces (see
[3]), “metric spaces” can be replaced by “nonzero Hausdorff topological vector spaces” in
definition3.4and in theorem3.5in which unboundedness of topological vector spaces need not
be mentioned. The following corollary is the one corresponding to the corollary 3.6, but for
linear mappings.
Corollary 3.8. Let f :X →X be a linear mapping on a nonzero Hausdorff topological vec-tor space X . Suppose, to each open neighbourhood U of 0 in X , there corresponds an open
Conflict of Interests
The authors declare that there is no conflict of interests.
REFERENCES
[1] M. A. Khamsi and W. A. Kirk, An introduction to metric spaces and fixed point theory, John-Wiley, New York, 2001.
